The standard deviations of x_i (i=1, 2, ldots | vedclass

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(B) Given that the standard deviations of $x_i$ and $y_i$ are $a$ and $b$ respectively,we have $a^2 = \frac{1}{10} \sum x_i^2 - \bar{x}^2$ and $b^2 =

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$\sqrt{a^2 + b^2 - \frac{c}{5}}$

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(B) Given that the standard deviations of $x_i$ and $y_i$ are $a$ and $b$ respectively,we have $a^2 = \frac{1}{10} \sum x_i^2 - \bar{x}^2$ and $b^2 = \frac{1}{10} \sum y_i^2 - \bar{y}^2$. We are given $\sum_{i=1}^{10} (x_i - \bar{x})(y_i - \bar{y}) = c$. Let $d_i = x_i - y_i$. The mean of $d_i$ is $\bar{d} = \bar{x} - \bar{y}$. The variance of $d_i$ is $\sigma_d^2 = \frac{1}{10} \sum (d_i - \bar{d})^2 = \frac{1}{10} \sum ((x_i - y_i) - (\bar{x} - \bar{y}))^2$. $\sigma_d^2 = \frac{1}{10} \sum ((x_i - \bar{x}) - (y_i - \bar{y}))^2 = \frac{1}{10} \sum (x_i - \bar{x})^2 + \frac{1}{10} \sum (y_i - \bar{y})^2 - \frac{2}{10} \sum (x_i - \bar{x})(y_i - \bar{y})$. Since $\frac{1}{10} \sum (x_i - \bar{x})^2 = a^2$ and $\frac{1}{10} \sum (y_i - \bar{y})^2 = b^2$,and $\sum (x_i - \bar{x})(y_i - \bar{y}) = c$,we get: $\sigma_d^2 = a^2 + b^2 - \frac{2c}{10} = a^2 + b^2 - \frac{c}{5}$. Therefore,the standard deviation is $\sqrt{a^2 + b^2 - \frac{c}{5}}$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$2$	$3$	$x$	$5$	$4$

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